I claim that the only solutions are $(m,n)=(3a,2b)$ and its permutations. For construction, use $ 3 \times 2$ rectangles repeteadly. Call the cells such that makes our any figure square, $f$-cells. Call $180^{\circ}$ rotated trominos of any tromino $friend$ of tromino. In order to prove our claim, it is enough to show that we must use $3 \times 2$ rectangles. Put any tromino to our rectangle. We must use some tromino to occupy $f$-cell. If it is not the $friend$ tromino, the other trominos we add will make $ 3 \times 3 $ square without $3$ cells, but we have to occupy these too, then using more trominos, we can't make $f$-cells $0$ since adding a tromino makes at least another $f$-cell, hence we must have $0$ $f$-cell from start, which is $3 \times 2$ square, hence conclusion.